INST 155 (Networks and Systems), section 1

Transcription

1 INST 155 (Networks and Systems), section 1 Recommended schedule Day 1 Day 2 Day 3 Day 4 Day 5 Day 6 Topics: Basic algebra review Questions: 1 through 10 Lab Exercise: Work on project (progress report: Question 61) Topics: Algebraic equation manipulation Questions: 11 through 20 Lab Exercise: Work on project (progress report: Question 62) Topics: Algebraic substitution Questions: 21 through 30 Lab Exercise: Work on project (progress report: Question 63) Topics: Solving simultaneous systems of equations Questions: 31 through 40 Lab Exercise: Work on project (progress report: Question 64) Topics: Resistor network analysis using simultaneous equations Questions: 41 through 50 Lab Exercise: Voltage divider with limited range (Question 65) Topics: Review Questions: 51 through 60 Lab Exercise: Work on project (progress report: Question 66) Impending deadlines Project due at end of INST155, Section 2 Question 67: Sample project grading criteria 1

2 INST 155 (Networks and Systems), section 1 Skill standards addressed by this course section This needs to be completed! EIA Raising the Standard; Electronics Technician Skills for Today and Tomorrow, June 1994 B Basic and Practical Skills Communicating on the Job B.01 Use effective written and other communication skills. Met by group discussion and completion of labwork. B.03 Employ appropriate skills for gathering and retaining information. Met by research and preparation prior to group discussion. B.04 Interpret written, graphic, and oral instructions. Met by completion of labwork. B.06 Use language appropriate to the situation. Met by group discussion and in explaining completed labwork. B.07 Participate in meetings in a positive and constructive manner. Met by group discussion. B.08 Use job-related terminology. Met by group discussion and in explaining completed labwork. B.10 Document work projects, procedures, tests, and equipment failures. Met by project construction and/or troubleshooting assessments. C Basic and Practical Skills Solving Problems and Critical Thinking C.01 Identify the problem. Met by research and preparation prior to group discussion. C.03 Identify available solutions and their impact including evaluating credibility of information, and locating information. Met by research and preparation prior to group discussion. C.07 Organize personal workloads. Met by daily labwork, preparatory research, and project management. C.08 Participate in brainstorming sessions to generate new ideas and solve problems. Met by group discussion. D Basic and Practical Skills Reading D.01 Read and apply various sources of technical information (e.g. manufacturer literature, codes, and regulations). Met by research and preparation prior to group discussion. E Basic and Practical Skills Proficiency in Mathematics E.01 Determine if a solution is reasonable. E.02 Demonstrate ability to use a simple electronic calculator. E.06 Translate written and/or verbal statements into mathematical expressions. E.07 Compare, compute, and solve problems involving binary, octal, decimal, and hexadecimal numbering systems. E.12 Interpret and use tables, charts, maps, and/or graphs. E.13 Identify patterns, note trends, and/or draw conclusions from tables, charts, maps, and/or graphs. E.15 Simplify and solve algebraic expressions and formulas. E.16 Select and use formulas appropriately. E.18 Use properties of exponents and logarithms. E.20 Graph functions. E.29 Graph basic functions using polar and/or Cartesian coordinate systems. 2

3 Question 1 When evaluating (calculating) a mathematical expression, what order should you do the various expressions in? In other words, which comes first: multiplication, division, addition, subtraction, powers, roots, parentheses, etc.; and then what comes after that, and after that? file Answer 1 Do what is inside parentheses first (the furthest inside parentheses if there are multiple layers of parentheses), powers and roots, functions (trig, log, etc.), multiplication/division, and finally addition/subtraction. Notes 1 Order of operations is extremely important, as it becomes critical to recognize proper order of evaluation when stripping an expression down to isolate a particular variable. In essence, the normal order of operations is reversed when undoing an expression, so students must recognize what the proper order of operations is. 3

4 Question 2 When evaluating an expression such as this, it is very important to follow proper order of operations. Otherwise, the correct result will be impossible to arrive at: 3log To show what the proper order of operations is for this expression, I show it being evaluated step by step here : 3log log Do the same for each of the following expressions: (3 + 11) 21 (7 4) 40 log ( ) 2 file Answer 2 I ll let you determine and document the proper order of operations, but here are the results of each expression: = = (3 + 11) = (7 4) 40 = log = ( ) 2 = By the way, this is a highly recommended practice for those struggling with mathematical principles: document each and every step by re-writing the expression. Although it takes more paper and more effort, it will save you from needless error and frustration! 4

5 Notes 2 Order of operations is extremely important, as it becomes critical to recognize proper order of evaluation when stripping an expression down to isolate a particular variable. In essence, the normal order of operations is reversed when undoing an expression, so students must recognize what the proper order of operations is. 5

6 Question 3 Observe the following equivalence: = (4 4 4) (4 4) Since all operations are the same (multiplication) and reversible, the parentheses are not needed. Therefore, we may write the expression like this: Of course, the simplest way to write this is 4 5, since there are five 4 s multiplied together. Expand each of these expressions so that there are no exponents either: = = = = After expanding each of these expressions, re-write each one in simplest form: one number to a power, just like the final form of the example given (4 5 ). From these examples, what pattern do you see with exponents of products. In other words, what is the general solution to the following expression? file Answer 3 a m a n = a m a n = a m+n Notes 3 I have found that students who cannot fathom the general rule (a m a n = a m+n ) often understand for the first time when they see concrete examples. 6

7 Question 4 Observe the following equivalence: = It should be readily apparent that we may cancel out two quantities from both top and bottom of the fraction, so in the end we are left with this: Re-writing this using exponents, we get Expand each of these expressions so that there are no exponents either: = = = = After expanding each of these expressions, re-write each one in simplest form: one number to a power, just like the final form of the example given (4 1 ). From these examples, what pattern do you see with exponents of products. In other words, what is the general solution to the following expression? file a m a n = Answer 4 a m a n = am n Notes 4 I have found that students who cannot fathom the general rule ( am a n first time when they see concrete examples. = a m n ) often understand for the 7

8 Question 5 Observe the following equivalence: = It should be readily apparent that we may cancel out two quantities from both top and bottom of the fraction, so in the end we are left with this: 1 4 Following the rule of am a = a m n, the reduction of 42 n 4 should be 4 1. Many students find this confusing, 3 as the intuitive concept of exponents (how many times a number is to be multiplied by itself) fails here. How in the world do we multiply 4 by itself -1 times?! Expand each of these expressions so that there are no exponents either: = = = = After expanding each of these expressions, re-write each one in simplest form: one number to a power, just like the final form of the example given (4 1 ), following the rule am a = a m n. From these examples, n what easy-to-understand definition can you think of to describe negative exponents? Also, expand the following expression so there are no exponents, then re-write the result in exponent form following the rule am a = a m n : n What does this tell you about exponents of zero? file Answer A negative exponent is simply the reciprocal (1/x) of its positive counterpart. A zero exponent is always equal to 1. Notes 5 I have found that students who cannot fathom the meaning of negative or zero exponents often understand immediately when they construct their own definition based on the general rule ( am a n = a m n ). 8

9 Question 6 The equation for calculating total resistance in a parallel circuit (for any number of parallel resistances) is sometimes written like this: R total = (R R R 1 n ) 1 Re-write this equation in such a way that it no longer contains any exponents. file Answer 6 1 R total = 1 R R 2 + R n Notes 6 This question is an exercise in basic algebra, specifically the meaning of negative exponents. 9

10 Question 7 A function is a mathematical relationship with an input (usually x) and an output (usually y). Here is an example of a simple function: y = 2x + 1 One way to show the pattern of any given function is with a table of numbers. Complete this table for the given values of x: x 2x A more common (and intuitive) way to show the pattern of any given function is with a graph. Complete this graph for the same function y = 2x + 1. Consider each division on the axes to be 1 unit: y x file

11 Answer 7 x 2x y x Notes 7 It is very important for your students to understand graphs, as they are very frequently used to illustrate the behavior of circuits and mathematical functions alike. Discuss with them how the line represents a continuous string of points and not just the integer values calculated in the table. 11

12 Question 8 Match each written function (y = ) with the sketched graph it fits best: y = 3x + 2 y = 5 2x y = x 2 y = 2 x file Answer 8 y = x 2 y = 3x + 2 y = 5-2x y = 2 x 12

13 Notes 8 The primary purpose of this question is to have students figure out how to match each expression to a graph. Of course, one could take the time to plot each function one by one, but there exist much simpler ways to determine the character of a function without plotting the whole thing. 13

14 Question 9 Many different equations used in the analysis of electric circuits may be graphed. Take for instance Ohm s Law for a 1 kω resistor: 1 kω 20 Current (I) in milliamps Voltage (V) in volts Plot this graph, following Ohm s Law. Then, plot another graph representing the voltage/current relationship of a 2 kω resistor. file

15 Answer kω Current (I) in milliamps 10 2 kω Voltage (V) in volts Notes 9 Ask your students to explain how they plotted the two functions. Did they make a table of values first? Did they draw dots on the paper and then connect those dots with a line? Did anyone plot dots for the endpoints and then draw a straight line in between because they knew this was a linear function? 15

16 Question 10 Many different equations used in the analysis of electric circuits may be graphed. Take for instance Ohm s Law for a variable resistor connected to a 12 volt source: 10 Current (I) in milliamps Resistance (R) in kilo-ohms Plot this graph, following Ohm s Law. file Answer Current (I) in milliamps Resistance (R) in kilo-ohms 16

17 Notes 10 Ask your students to explain how they plotted the two functions. Did they make a table of values first? Did they draw dots on the paper and then connect those dots with a line? Did anyone plot dots for the endpoints and then draw a straight line in between because they knew this was a linear function? Many students are surprised that the plot is nonlinear, being that resistors are considered linear devices! 17

18 Question 11 Solve for n in the following equations: Equation 1: 56 = 14n Equation 2: 54 n = 10 Equation 3: 4 n = 12 Equation 4: 28 = 2 n file Answer 11 Notes 11 Equation 1: n = 4 Equation 2: n = 44 Equation 3: n = Equation 4: n = 26 Have your students come to the front of the class and show everyone else the techniques they used to solve for the value of a in each equation. Remind them to document each and every step in the process, so that nothing is left to guess or to chance. Equations 2 through 4 require two steps to solve for n. Equation 1 only requires a single step, but the two negative numbers may be a bit confusing to some. 18

19 Question 12 The formula for calculating total resistance of three series-connected resistors is as follows: R = R 1 + R 2 + R 3 Algebraically manipulate this equation to solve for one of the series resistances (R 1 ) in terms of the other two series resistances (R 2 and R 3 ) and the total resistance (R). In other words, write a formula that solves for R 1 in terms of all the other variables. file Answer 12 R 1 = R (R 2 + R 3 ) or R 1 = R R 2 R 3 Notes 12 This question is nothing more than practice algebraically manipulating equations. Ask your students to show you how they solved it, and how the two given answers are equivalent. 19

20 Question 13 Manipulate this equation to solve for resistor value R 1, given the values of R 2 and R parallel : R parallel = R 1R 2 R 1 + R 2 Then, give an example of a practical situation where you might use this new equation. file Answer 13 Notes 13 R 1 = R 2R parallel R 2 R parallel I ll let you figure out a situation where this equation would be useful! This question is really nothing more than an exercise in algebraic manipulation. 20

21 Question 14 The formula for calculating total resistance of three parallel-connected resistors is as follows: 1 R = 1 R R R 3 Algebraically manipulate this equation to solve for one of the parallel resistances (R 1 ) in terms of the other two parallel resistances (R 2 and R 3 ) and the total resistance (R). In other words, write a formula that solves for R 1 in terms of all the other variables. file Answer 14 1 R 1 = 1 R ( 1 R R 3 ) 1 or R 1 = 1 R 1 R 2 1 R 3 Notes 14 This question is nothing more than practice algebraically manipulating equations. Ask your students to show you how they solved it, and how the two given answers are equivalent. 21

22 Question 15 The following equations solve for the output voltage of various switching converter circuits, given the switch duty cycle D and the input voltage: V out = D V in (Buck converter circuit) V out = V in 1 D (Boost converter circuit) V out = D V in 1 D (Inverting or Cuk converter circuit) Manipulate each of these equations to solve for duty cycle (D) in terms of the input voltage (V in ) and desired output voltage (V out ). Remember that duty cycle is always a quantity between 0 and 1, inclusive. file Answer 15 D = V out V in ( ) Vin D = 1 V out (Buck converter circuit) (Boost converter circuit) V out D = V in + V out (Inverting or Cuk converter circuit) Notes 15 Given the equations for these converter circuit types solving for output voltage in terms of input voltage and duty cycle D, this question is nothing more than an exercise in algebraic manipulation. Note to your students that all of these equations are subject to conditions of loading. That is to say, variations in load will skew the output voltage as as predicted by these neat, simple formulae. Although these DC-DC power converter circuits are commonly referred to as regulators, it is somewhat misleading to do so because it falsely implies a capacity for self-correction of output voltage. Only when coupled to a feedback control network are any of these converter circuits capable of actually regulating output voltage to a set value. 22

23 Question 16 The power dissipation of a transistor is given by the following equation: ( P = I C V CE + V ) BE β Manipulate this equation to solve for beta, given all the other variables. file Answer 16 β = V BE P I C V CE Notes 16 Although this question is essentially nothing more than an exercise in algebraic manipulation, it is also a good lead-in to a discussion on the importance of power dissipation as a semiconductor device rating. High temperature is the bane of most semiconductors, and high temperature is caused by excessive power dissipation. A classic example of this, though a bit dated, is the temperature sensitivity of the original germanium transistors. These devices were extremely sensitive to heat, and would fail rather quickly if allowed to overheat. Solid state design engineers had to be very careful in the techniques they used for transistor circuits to ensure their sensitive germanium transistors would not suffer from thermal runaway and destroy themselves. Silicon is much more forgiving then germanium, but heat is still a problem with these devices. At the time of this writing (2004), there is promising developmental work on silicon carbide and gallium nitride transistor technology, which is able to function under much higher temperatures than silicon. 23

24 + Question 17 The equation for voltage gain (A V ) in a typical inverting, single-ended opamp circuit is as follows: A V = R 1 R 2 Where, R 1 is the feedback resistor (connecting the output to the inverting input) R 2 is the other resistor (connecting the inverting input to voltage signal input terminal) Suppose we wished to change the voltage gain in the following circuit from 3.5 to 4.9, but only had the freedom to alter the resistance of R 2 : V out 7k7 R 1 2k2 R 2 V in Algebraically manipulate the gain equation to solve for R 2, then determine the necessary value of R 2 in this circuit to give it a voltage gain of 4.9. file Answer 17 Notes 17 R 2 = R 1 A V For the circuit shown, R 2 would have to be set equal to kω. Nothing more than a little algebra to obtain the answers for this question! 24

25 Question 18 The equation for voltage gain (A V ) in a typical noninverting, single-ended opamp circuit is as follows: A V = R 1 R Where, R 1 is the feedback resistor (connecting the output to the inverting input) R 2 is the other resistor (connecting the inverting input to ground) Suppose we wished to change the voltage gain in the following circuit from 5 to 6.8, but only had the freedom to alter the resistance of R 2 : V in R 2 1k175 + R 1 4k7 V out Algebraically manipulate the gain equation to solve for R 2, then determine the necessary value of R 2 in this circuit to give it a voltage gain of 6.8. file Answer 18 Notes 18 R 2 = R 1 A V 1 For the circuit shown, R 2 would have to be set equal to Ω. Nothing more than a little algebra to obtain the answers for this question! 25

26 Question 19 The decay of a variable over time in an RC or LR circuit follows this mathematical expression: Where, e = Euler s constant ( ) t = Time, in seconds τ = Time constant of circuit, in seconds e t τ For example, if we were to evaluate this expression and arrive at a value of 0.398, we would know the variable in question has decayed from 100% to 39.8% over the period of time specified. However, calculating the amount of time it takes for a decaying variable to reach a specified percentage is more difficult. We would have to manipulate the equation to solve for t, which is part of an exponent. Show how the following equation could be algebraically manipulated to solve for t, where x is the number between 0 and 1 (inclusive) representing the percentage of original value for the variable in question: x = e t τ Note: the trick here is how to isolate the exponent t τ. You will have to use the natural logarithm function! file Answer 19 Showing all the necessary steps: x = e t τ ( ) lnx = ln e t τ lnx = t τ t = τ lnx 26

27 Notes 19 In my experience, most American high school graduates are extremely weak in logarithms. Apparently this is not taught very well at the high school level, which is a shame because logarithms are a powerful mathematical tool. You may find it necessary to explain to your students what a logarithm is, and exactly why it un-does the exponent. When forced to give a quick presentation on logarithms, I usually start with a generic definition: Given: b a = c Logarithm defined: log b c = a Verbally defined, the logarithm function asks us to find the power (a) of the base (b) that will yield c. Next, I introduce the common logarithm. This, of course, is a logarithm with a base of 10. A few quick calculator exercises help students grasp what the common logarithm function is all about: log 10 = log 100 = log 1000 = log = log = log 1 10 = log = log = After this, I introduce the natural logarithm: a logarithm with a base of e (Euler s constant): Natural logarithm defined: lnx = log e x Have your students do this simple calculation on their calculators, and explain the result: ln = Next comes an exercise to help them understand how logarithms can undo exponentiation. Have your students calculate the following values: e 2 = e 3 = e 4 = 27

28 Now, have them take the natural logarithms of each of those answers. They will find that they arrive at the original exponent values (2, 3, and 4, respectively). Write this relationship on the board as such for your students to view: lne 2 = 2 lne 3 = 3 lne 4 = 4 Ask your students to express this relationship in general form, using the variable x for the power instead of an actual number: lne x = x It should now be apparent that the natural logarithm function has the ability to undo a power of e. Now it should be clear to your students why the given sequence of algebraic manipulations in the answer for this question is true. 28

29 Question 20 The electrical resistance of a conductor at any temperature may be calculated by the following equation: R T = R r + R r αt R r αt r Where, R T = Resistance of conductor at temperature T R r = Resistance of conductor at reference temperature T r α = Temperature coefficient of resistance at reference temperature T r Simplify this equation by means of factoring. file Answer 20 R T = R r [1 + α(t T r )] Follow-up question: when plotted on a graph with temperature (T) as the independent variable and resistance (R T ) as the dependent variable (i.e. a two-axis graph with T on the horizontal and R on the vertical), is the resulting plot linear? Why or why not? How is it possible to tell just by looking at the equation, prior to actually plotting on a graph? Notes 20 Just an exercise in algebra here! 29

30 Question 21 Substitution is a technique whereby we let a variable represent (stand in the place of) another variable or an expression made of other variables. One application where we might use substitution is when we must manipulate an algebraic expression containing a lot of similar-looking variables, as is often the case with science problems. Take this series-parallel resistor circuit for example: R 1 R total R 2 R 3 R 4 The equation expressing total resistance as a function of the four resistor values looks like this: R total = R 1 + R 2(R 3 + R 4 ) R 2 + R 3 + R 4 Now imagine being asked to manipulate this equation to solve for R 3. When the only visual feature distinguishing each of the variables is the subscript (total, 1, 2, 3, or 4), it becomes very easy to lose track of where one is in the algebraic manipulation. A very common mistake is to exchange or needlessly repeat subscripts during the process, effectively mis-placing one or more variables. To help avoid such mistakes, you may substitute different letter variables for R total, R 1, R 2, R 3, and R 4 like this: Substitution table Original variable R total R 1 R 2 R 3 R 4 New variable y a b c d y = a + b(c + d) b + c + d After doing the algebraic manipulation to solve for c (R 3 ), the equation looks like this: c = (y a)(b + d) bd a + b y Back-substitute the original R variables in place of a, b, c, d, and y as you see them in the above equation to arrive at a form that directly relates to the schematic diagram. file

31 Answer 21 R 3 = (R total R 1 )(R 2 + R 4 ) R 2 R 4 R 1 + R 2 R total Notes 21 Challenge question: show all the steps you would take to solve for R 3 in the original equation. Here I show an application of substitution that is useful only because the human brain has difficulty distinguishing similar-looking symbols. More powerful uses of algebraic substitution exist, of course, but this is a start for students who have never seen the concept before. 31

32 Question 22 Substitution is a technique whereby we let a variable represent (stand in the place of) another variable or an expression made of other variables. One application where we might use substitution is when we must manipulate an algebraic expression containing multiple instances of the same sub-expression. For example, suppose we needed to manipulate this equation to solve for c: 1 = a + b(d2 f 2 ) + c d 2 f 2 The sub-expression d 2 f 2 appears twice in this equation. Wouldn t it be nice if we had something simpler to put in its place during the time we were busy manipulating the equation, if for no other reason than to have less variables to write on our paper while showing all the steps to our work? Well, we can do this! Substitute the variable x for the sub-expression d 2 f 2, and then solve for c. When you are done manipulating the equation, back-substitute d 2 f 2 in place of x. file Answer 22 Original equation: 1 = a + b(d2 f 2 ) + c d 2 f 2 After substituting x: 1 = a + bx + c x After manipulating the equation to solve for c: c = x(1 b) a Back-substituting the original sub-expression in place of x: c = (d 2 f 2 )(1 b) a Notes 22 Here I show an application of substitution that is useful only because the human brain has an easier time dealing with a single symbol than with a collection of different symbols. More powerful uses of algebraic substitution exist, of course, but this is a start for students who are new to the concept. 32

33 Question 23 Substitution is the term we give to the mathematical equivalence of one variable to one or more other variables in an expression. It is a fundamental principle used to combine two or more equations into a single equation (among other things). For example, we know that the formula for calculating current in a simple one-resistor circuit is as follows: V R I = V / R We also know that the total resistance (R) of a three-resistor series circuit is as follows: R 1 R = R 1 + R 2 + R 3 R 2 R 3 Combine these two equations together using substitution so that we have a single equation for calculating current I in a three-resistor series circuit given the source voltage V and each resistance value R 1, R 2, and R 3 : R 1 V R 2 R 3 I = f(v, R 1, R 2, R 3 ) "I is some function of V, R 1, R 2, and R 3 " In other words, you need to have as your answer a single equation that begins with I = and has all the variables V, R 1, R 2, and R 3 on the other side of the equal sign. 33

34 file Answer 23 V I = R 1 + R 2 + R 3 Notes 23 I like to speak of the process of substitution in terms of definitions for variables. In this particular case, R 1 + R 2 + R 3 is a definition for R that we put in R s place in the first equation (I = V R ). The notation shown in the third schematic, I = f(v,r 1,R 2,R 3 ), is known as function notation. It merely means that the value of I is determined by the values of all those variables within the parentheses, rather than just one. 34

35 Question 24 We know that the current in a series circuit may be calculated with this formula: I = E total R total We also know that the voltage dropped across any single resistor in a series circuit may be calculated with this formula: E R = IR Combine these two formulae into one, in such a way that the I variable is eliminated, leaving only E R expressed in terms of E total, R total, and R. file Answer 24 E R = E total ( R R total ) Follow-up question: algebraically manipulate this equation to solve for E total in terms of all the other variables. In other words, show how you could calculate for the amount of total voltage necessary to produce a specified voltage drop (E R ) across a specified resistor (R), given the total circuit resistance (R total ). Notes 24 Though this voltage divider formula may be found in any number of electronics reference books, your students need to understand how to algebraically manipulate the given formulae to arrive at this one. 35

36 Question 25 We know that the voltage in a parallel circuit may be calculated with this formula: E = I total R total We also know that the current through any single resistor in a parallel circuit may be calculated with this formula: I R = E R Combine these two formulae into one, in such a way that the E variable is eliminated, leaving only I R expressed in terms of I total, R total, and R. file Answer 25 Notes 25 ( Rtotal ) I R = I total R How is this formula similar, and how is it different, from the voltage divider formula? Though this current divider formula may be found in any number of electronics reference books, your students need to understand how to algebraically manipulate the given formulae to arrive at this one. At first it may seem as though the two divider formulae (voltage versus current) are easy to confuse. R R total Is it or R total R? However, there is a very simple way to remember which fraction belongs with which formula, based on the numerical value of that fraction. Mention this to your students and at least one of them will be sure to recognize the pattern. 36

37 Question 26 There are two basic Ohm s Law equations: one relating voltage, current, and resistance; and the other relating voltage, current, and power (the latter equation is sometimes known as Joule s Law rather than Ohm s Law): E = IR P = IE In electronics textbooks and reference books, you will find twelve different variations of these two equations, one solving for each variable in terms of a unique pair of two other variables. However, you need not memorize all twelve equations if you have the ability to algebraically manipulate the two simple equations shown above. Demonstrate how algebra is used to derive the ten other forms of the two Ohm s Law / Joule s Law equations shown here. file

38 Answer 26 I won t show you how to do the algebraic manipulations, but I will show you the ten other equations. First, those equations that may be derived strictly from E = IR: I = E R R = E I Next, those equations that may be derived strictly from P = IE: I = P E E = P I Next, those equations that may be derived by using algebraic substitution between the original two equations given in the question: P = I 2 R P = E2 R And finally, those equations which may be derived from manipulating the last two power equations: R = P I 2 I = P R E = PR R = E2 P Notes 26 Algebra is an extremely important tool in many technical fields. One nice thing about the study of electronics is that it provides a relatively simple context in which fundamental algebraic principles may be learned (or at least illuminated). The same may be said for calculus concepts as well: basic principles of derivative and integral (with respect to time) may be easily applied to capacitor and inductor circuits, providing students with an accessible context in which these otherwise abstract concepts may be grasped. But calculus is a topic for later worksheet questions... 38

39 Question 27 Suppose we only knew the emitter and base currents for an operating transistor and wished to calculate β from that information. We would need a definition of beta cast in terms of I E and I B instead of I C and I B. Apply algebraic substitution to the formula β = IC I B so that beta (β) is defined in terms of I E and I B. You may find the following equation helpful in your work: file Answer 27 I E = I C + I B β = I E I B 1 Notes 27 This question is nothing more than an exercise in algebraic manipulation. 39

40 Question 28 The resistance of a piece of copper wire at temperature T (in degrees Celsius) is given by the following formula: R T = R o [ (T 20)] Suppose you wished to alter this formula so it could accept values for T in units of degrees Fahrenheit instead of degrees Celsius. Suppose also that the only formula you are able to find for converting between Fahrenheit (T F ) and Celsius (T C ) is this one: ( ) 9 T F = T C Combined these two formulae into one solving for the resistance of a copper wire sample (R T ) at a specific temperature in degrees Celsius (T C ), given the specimen s reference resistance (R o ) at 20 o Celsius (room temperature). file Answer 28 ( )] 5 R T = R o [ T F Notes 28 Solving this algebraic problem requires both manipulation of the temperature equation and substitution of variables. One important detail I incorporated into this question is the lack of a subscript for T in the original resistance formula. In the first sentence I identify that temperature as being in degrees Celsius, but since there is no other T variables in the equation, I did not have to include a C subscript. When students look to the Celsius-Fahrenheit conversion formula to substitute into the resistance formula, they must decide which T in the conversion formula to use, T F or T C. Here, I purposely wrote the conversion formula in terms of T F to see how many students would blindly substitute T F for T in the resistance formula instead of properly identifying T C as the variable to substitute and doing the work of manipulation. Far from being a trick question, this scenario is very realistic. Formulae found in reference manuals do not necessarily use standardized variables, but rather cast their variables according to context. Multiple formulae will most likely not be written with identical subscripted variables just waiting to be substituted. It is the domain of the intelligent technician, engineer, or scientist to figure out what variables are appropriate to substitute based on context! 40

41 Question 29 A bipolar junction transistor parameter similar to β is alpha, symbolized by the Greek letter α. It is defined as the ratio between collector current and emitter current: α = I C I E Apply algebraic substitution to this formula so that alpha is defined as a function of beta: α = f(β). In other words, substitute and manipulate this equation until you have alpha by itself on one side and no variable except beta on the other. You may find the following equations helpful in your work: file Answer 29 β = I C I B α = β β + 1 I E = I C + I B Notes 29 Follow-up question: what range of values might you expect for α, with a typical transistor? This question is nothing more than an exercise in algebraic manipulation. 41

42 Question 30 The Q factor of a series inductive circuit is given by the following equation: Q = X L R series Likewise, we know that inductive reactance may be found by the following equation: X L = 2πfL We also know that the resonant frequency of a series LC circuit is given by this equation: 1 f r = 2π LC Through algebraic substitution, write an equation that gives the Q factor of a series resonant LC circuit exclusively in terms of L, C, and R, without reference to reactance (X) or frequency (f). file Answer 30 Q = 1 R L C Notes 30 This is merely an exercise in algebra. However, knowing how these three component values affects the Q factor of a resonant circuit is a valuable and practical insight! 42

43 Question 31 Solve this equation for the value of x: x + 5 = 8 How many exact solutions does the above equation have? Now, determine a few different solutions for the following equation: x + y = 8 How many exact solutions does this equation have? Plot this equation s solutions on the following graph: y x file

44 Answer 31 If x + 5 = 8, then x = 3 (exactly one solution) If x + y = 8, then there are an infinite number of solutions: y x + y = 8 x Notes 31 Follow-up question: find the solution to x + 5 = 8 on this same graph. This question begins with an extremely simple equation having one solution and moves on to another simple equation having an infinite number of solutions. While an infinitude of correct answers may seem impossible to rationally deal with, a graph handles it quite nicely, the multitude of correct answer pairs represented as a line on a graph with infinite length. 44

45 Question 32 Plot the solutions to the equation y + x = 8 on a graph: y x On the same graph, plot the solutions to the equation y x = 3. What is the significance of the point where the two lines cross? file Answer 32 y y + x = 8 y - x = 3 (2.5, 5.5) x The point of intersection between the two lines represents the one solution set that satisfies both equations (where x = 2.5 and y = 5.5). Notes 32 The purpose of this question is to gently introduce students to the concept of simultaneous systems of equations, where a set of solutions satisfies more than one equation at a time. It is important for students to understand the basic concepts of graphing before they try to answer this question, though. 45

46 Question 33 What does it actually mean to obtain a solution for a simultaneous system of equations? For example, if 2x + y = 7 and x y = 1, what do the solution values (x = 2 ; y = 3) represent? If we were to graph both these linear equations on a Cartesian (x, y) coordinate system, where would the solution (2,3) be located on the graph? file Answer 33 The solutions for a system of equations represent a unique combination of values that satisfy all equations in that system. For a two variable system, the solution is the intersection of two lines. Notes 33 Many students have difficulty grasping the significance of systems of equations. Discuss the meaning of equations, and systems of equations, with your students, making sure that the concept of simultaneity (solutions satisfying all equations at once) is made clear. 46

47 Question 34 Plot the equation y = x 2 on the following graph: y x On the same graph, plot the equation y = x + 2. What is the significance of the point where the two plots cross? file Answer 34 y y = x 2 y = x + 2 (-1, 1) (2, 4) x Here there are two points of intersection between the parabola (curve) and the straight line, representing two different solution sets that satisfy both equations. Challenge question: solve this simultaneous system of equations without graphing, but by symbolically manipulating the equations! 47

48 Notes 34 Here, solution by graphing may be a bit easier than the symbolic solution. In principle we may determine solutions for any pair of equations by graphing, with about equal difficulty. The only real problem is precision: how closely we may interpret to points of intersection. A practical example of non-linear simultaneous function solution is load line analysis in semiconductor circuitry. 48

49 Question 35 Load lines are useful tools for analyzing transistor amplifier circuits, but they may be hard to understand at first. To help you understand what load lines are useful for and how they are determined, I will apply one to this simple two-resistor circuit: Load 1.5 kω R 1 A 1 kω 20 V B We will have to plot a load line for this simple two-resistor circuit along with the characteristic curve for resistor R 1 in order to see the benefit of a load line. Load lines really only have meaning when superimposed with other plots. First, the characteristic curve for R 1, defined as the voltage/current relationship between terminals A and B: Characteristic "curve" for R 1 I R1 (ma) V AB Next, I will plot the load line as defined by the 1.5 kω load resistor. This load line expresses the voltage available between the same two terminals (V AB ) as a function of the load current, to account for voltage dropped across the load: 49

50 I R1 (ma) Characteristic "curve" for R 1 Load line V AB At what value of current (I R1 ) do the two lines intersect? Explain what is significant about this value of current. file Answer 35 I R = 8 ma is the same value of current you would calculate if you had analyzed this circuit as a simple series resistor network. Follow-up question: you might be wondering, what is the point of plotting a characteristic curve and a load line in such a simple circuit, if all we had to do to solve for current was add the two resistances and divide that total resistance value into the total voltage? Well, to be honest, there is no point in analyzing such a simple circuit in this manner, except to illustrate how load lines work. My follow-up question to you is this: where would plotting a load line actually be helpful in analyzing circuit behavior? Can you think of any modifications to this two-resistor circuit that would require load line analysis in order to solve for current? Notes 35 While this approach to circuit analysis may seem silly using load lines to calculate the current in a two-resistor circuit it demonstrates the principle of load lines in a context that should be obvious to students at this point in their study. Discuss with your students how the two lines are obtained (one for resistor R 1 and the other plotting the voltage available to R 1 based on the total source voltage and the load resistor s value). Also, discuss the significance of the two line intersecting. Mathematically, what does the intersection of two graphs mean? What do the coordinate values of the intersection point represent in a system of simultaneous functions? How does this principle relate to an electronic circuit? 50

51 Question 36 Load lines are useful tools for analyzing transistor amplifier circuits, but they may be applied to other types of circuits as well. Take for instance this diode-resistor circuit: Load D 1 A B 2.5 kω 5 V The diode s characteristic curve is already plotted on the following graph. Your task is to plot the load line for the circuit on the same graph, and note where the two lines intersect: I D (ma) V AB What is the practical significance of these two plots intersection? file

52 Answer 36 The two lines intersect at a current of approximately 1.72 ma: I D (ma) V AB Follow-up question: explain why the use of a load line greatly simplifies the determination of circuit current in such a diode-resistor circuit. Challenge question: suppose the resistor value were increased from 2.5 kω to 10 kω. What difference would this make in the load line plot, and in the intersection point between the two plots? Notes 36 While this approach to circuit analysis may seem silly using load lines to calculate the current in a diode-resistor circuit it demonstrates the principle of load lines in a context that should be obvious to students at this point in their study. Discuss with your students how the load line is obtained for this circuit, and why it is straight while the diode s characteristic curve is not. Also, discuss the significance of the two line intersecting. Mathematically, what does the intersection of two graphs mean? What do the coordinate values of the intersection point represent in a system of simultaneous functions? How does this principle relate to an electronic circuit? 52

53 Question 37 Suppose you were given the following two equations and asked to find solutions for x and y that will satisfy both at the same time: y + x = 8 y x = 3 If we manipulate the second equation so as to solve for y, we will have a definition of y in terms of x that we may use for substitution in the first equation: y = x + 3 Show the process of substitution into the first equation, and how this leads to a single solution for x. Then, use that value of x to solve for y, resulting in a solution set valid for both equations. file Answer 37 If y + x = 8 and y = x + 3, then (x + 3) + x = 8. Therefore, x = 2.5 and y = 5.5 Notes 37 This question demonstrates one of the (many) practical uses of algebraic substitution: simultaneous systems of equations. solving 53

54 Question 38 Suppose you were given the following two equations and asked to find solutions for x and y that will satisfy both at the same time: y + x = 8 y x = 3 Now, you know that we may do anything we want to either equation as long as we do the same thing to both sides (on either side of the equal sign). This is the basic rule we follow when manipulating an equation to solve for a particular variable. For example, we may take the equation y + x = 8 and subtract x from both sides to yield an equation expressed in terms of y: y + x = 8 -x -x y = 8 - x Following the same principle, we may take two equations and combine them either by adding or subtracting both sides. For example, we may take the equation y x = 3 and add both sides of it to the respective sides of the first equation y + x = 8: y + x = 8 y - x = 3 2y = 11 What beneficial result comes of this action? In other words, how can I use this new equation 2y = 11 to solve for values of x and y that satisfy both of the original equations? file Answer 38 We may use the result (2y = 11) to solve for a value of y, which when substituted into either of the original equations may be used to solve for a value of x to satisfy both equations at the same time. Notes 38 While not intuitively obvious to most people, the technique of adding two entire equations to each other for the purpose of eliminating a variable is not only possible to do, but very powerful when looking for solutions to satisfy both original equations. Discuss with your students why it is allowable for us to add y x to y + x and to add 3 to 8. to yield the equation 2y =

55 Question 39 Solve for values of x and y that will satisfy both of the following equations at the same time: x + 2y = 9 file Answer 39 x = 3 y = 6 4x y = 18 Follow-up question: solve this system of simultaneous equations using both substitution (solving for one variable in one of the equations and substituting that into the other equation) and addition (adding the two equations together to produce a third equation with only one unknown). Notes 39 Nothing special here just practice solving for a two-variable system of equations. 55

56 Question 40 Many circuit analysis techniques require the solution of systems of linear equations, sometimes called simultaneous equations. This question is really a series of practice problems for solving simultaneous linear equations, the purpose being to give you lots of practice using various solution techniques (including the solution facilities of your calculator). Systems of two variables: x + y = 5 x y = 6 2x + y = 7 x y = 1 2x y = 4 x y = 2 3x 2y = 1 10x + 2y = 0 3x 5y = 13 5x + y = 6 3x 5y = 28 x + 2y = x 500y = x y = x 5000y = x y = x 2800y = x 2700y = 6.5 Systems of three variables: x y + z = 1 3x + 2y 5z = 21 x + y + z = 0 x y + z = 1 x 3y + z = 8 2x y 4z = 9 x + y + z = 3 x y z = 12 2x + 2y z = 12 x + y 2z = 12 4x 3y + 2z = 32 19x 6y + 20z = 33 3x 2y + z = 19 x 2y + 3z = 1 4x + 5y 3z = 17 4x + 3y 5z = 45 2x + 7y z = 3 7x + 2y 8z = 9 890x 1000y z = x 6200y z = x y 5100z = x y 1000z = 8100 x + y z = 0 6x 2y 3z = 5 file

57 Answer 40 Systems of two variables: x + y = 5 x y = 6 2x + y = 7 x y = 1 2x y = 4 x y = 2 x = 3 ; y = 2 x = 10 ; y = 16 x = 3 ; y = 1 3x 2y = 1 10x + 2y = 0 3x 5y = 13 5x + y = 6 3x 5y = 28 x + 2y = 5 x = 1 ; y = 1 x = 1 ; y = 5 x = 1 ; y = x 500y = x y = x 5000y = x y = x 2800y = x 2700y = 6.5 x = 1 ; y = 2 x = 5 ; y = 4 x = ; y = Systems of three variables: x y + z = 1 3x + 2y 5z = 21 x + y + z = 0 x y + z = 1 x 3y + z = 8 2x y 4z = 9 x + y + z = 3 x y z = 12 2x + 2y z = 12 x = 1 ; y = 1 ; z = 1 x = 4 ; y = 1 ; z = 7 x = 3 ; y = 3 ; z = 0 x + y 2z = 12 4x 3y + 2z = 32 19x 6y + 20z = 33 3x 2y + z = 19 x 2y + 3z = 1 4x + 5y 3z = 17 4x + 3y 5z = 45 2x + 7y z = 3 7x + 2y 8z = 9 x = 2 ; y = 4 ; z = 5 x = 6 ; y = 2 ; z = 1 x = 5 ; y = 3 ; z = 4 890x 1000y z = x 6200y z = x y 5100z = x y 1000z = 8100 x + y z = 0 6x 2y 3z = 5 x = ; y = ; z = x = ; y = ; z = Notes 40 I suggest you let your students discover how to use the equation-solving facilities of their scientific calculators on their own. My experience has been that students both young and old take to this challenge readily, because they realize learning how to use their calculators will save them a tremendous amount of hand calculations! 57

58 Question 41 Suppose you needed to choose a fixed resistor value (R) to make a voltage divider circuit, given a known potentiometer resistance value, the source voltage value, and the desired range of adjustment: 24 V R 50 kω V out Desired range for V out = 0 to 17 volts Solve for R, and show the equation you set up in order to do it. Hint: remember the series resistor voltage divider formula... ( ) R V R = V total R total file Answer 41 Notes 41 R = kω Be sure to have your students set up their equations in front of the class so everyone can see how they did it. Some students may opt to apply Ohm s Law to the solution of R, which is good, but for the purpose of developing equations to fit problems it might not be the best solution. Challenge your students to come up with a single equation that solves for R, with all known quantities on the other side of the equal sign. 58

59 Question 42 Suppose you needed to choose a potentiometer value (R) to make a voltage divider circuit, given a known fixed resistor value, the source voltage value, and the desired range of adjustment: 24 V 50 kω R V out Desired range for V out = 0 to 17 volts Solve for R, and show the equation you set up in order to do it. Hint: remember the series resistor voltage divider formula... ( ) R V R = V total R total file Answer 42 R = kω Follow-up question: you will not be able to find a potentiometer with a full-range resistance value of exactly kω. Describe how you could take a standard-value potentiometer and connect it to one or more fixed-value resistors to give it this desired full-scale range. Notes 42 Be sure to have your students set up their equations in front of the class so everyone can see how they did it. Some students may opt to apply Ohm s Law to the solution of R, which is good, but for the purpose of developing equations to fit problems it might not be the best solution. Challenge your students to come up with a single equation that solves for R, with all known quantities on the other side of the equal sign. The follow-up question is very practical, as it is impossible to find potentiometers ready-made to arbitrary values of full-scale resistance. Instead, you must work with what you can find, which is usually nominal values such as 10 kω, 100 kω, 1 MΩ, etc. 59